1. Heat flux through a 50 mm thick plane wall is q=70W/m^2 Determine the temperature difference at the wall surfaces and the numerical value of the temperature gradient in the wall if it is made of red bricks [lambda =0.70 W/(mcdot K)]

To determine the temperature difference at the wall surfaces and the numerical value of the temperature gradient in the wall, we can use the formula for heat flux through a plane wall:<br /><br />q = -λ * (dT/dx)<br /><br />where q is the heat flux, λ is the thermal conductivity of the wall material, and dT/dx is the temperature gradient in the wall.<br /><br />Given that the heat flux is q = 70 W/m^2 and the thermal conductivity of red bricks is λ = 0.70 W/(m·K), we can rearrange the formula to solve for the temperature gradient:<br /><br />dT/dx = -q/λ<br /><br />Substituting the given values, we get:<br /><br />dT/dx = -70 W/m^2 / 0.70 W/(m·K) = -100 K/m<br /><br />The negative sign indicates that the temperature decreases in the direction of the heat flux. Therefore, the numerical value of the temperature gradient in the wall is 100 K/m.<br /><br />To determine the temperature difference at the wall surfaces, we need to integrate the temperature gradient over the thickness of the wall. Assuming that the temperature gradient is constant throughout the wall, we can use the formula:<br /><br />ΔT = (dT/dx) * x<br /><br />where ΔT is the temperature difference and x is the thickness of the wall.<br /><br />Given that the thickness of the wall is x = 50 mm = 0.05 m, we can substitute the values into the formula:<br /><br />ΔT = (-100 K/m) * 0.05 m = -5 K<br /><br />The negative sign indicates that the temperature decreases in the direction of the heat flux. Therefore, the temperature difference at the wall surfaces is 5 K.